Find the centre of mass of three particles at the verticles of an equilateral triangle. The masses of the particles are $100 g$ , $150 g$ , and $200 g$ respectively. Each side of the equilateral triangle is $0.5 m$ long.

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Solution

With the $x$ - and $y$ - axes chosen as shown in Fig., the coordinates of points O, A and B forming the equilateral triangle are respectively $(0, 0), (0.5, 0), (0.25, 0.25 \sqrt{3})$ . Let the masses $100 g, 150 g$ and $200 g$ be located at O, A and B be respectively. Then,
$X = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$
$= \frac{100 (0) + 150 (0.5) + 200 (0.25) g m}{(100 + 150 + 200) g}$
$= \frac{75 + 50}{450} m = \frac{125}{450} m = \frac{5}{18} m$
$Y = \frac{100 (0) + 150 (0) + 200 (0.25 \sqrt{3}) g m}{450 g}$
$= \frac{50 \sqrt{3}}{450} m$ $= \frac{\sqrt{3}}{9} m = \frac{1}{3 \sqrt{3}} m$
The centre of mass C is shown in the figure. Note that it is not the geometric centre of the triangle OAB.

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